# Taylor's Theorem

• Apr 4th 2011, 12:47 PM
worc3247
Taylor's Theorem
Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is such that both $f\prime$ and $f\prime\prime$ exist for all $x \in \mathbb{R}$, so that Taylor's theorem tells us that, for each $a, h\in \mathbb{R}$ there is a $\theta$ such that $0<\theta<1$ so that $f(a+h)=f(a)+hf\prime(a)+\frac{1}{2}h^2 f\prime\prime(a+\theta h)$. Suppose further that on [0,2] the inequalities $|f(x)|\leq 1$ and $|f\prime\prime(x)| \leq 1$ hold. Write down the Taylor expansions of f(0) and f(2) about the point $x \in [0,2]$, using the above form of Taylor's Theorem, with a remainder involving $f\prime\prime$. Hence prove that for all $x \in [0,2]$ we have $|f\prime(x)| \leq 2$.

I'm not sure about the first part, what I have is:
$f(0)=f(-h)+hf\prime(-h)+\frac{1}{2}h^2f\prime\prime(h(\theta-1))$ by setting a+h=0 and
$f(2)=f(2-h)+hf\prime(2-h)+\frac{1}{2}f\prime\prime(2+h(\theta-1))$ but I don't think this is correct. It says about the point x in [0,2] but I don't know how to apply this to the formula. I have no idea for the last part. Thanks!
• Apr 5th 2011, 03:54 PM
choovuck
ignore this
• Apr 5th 2011, 03:55 PM
choovuck
Taylor's expansion about x is (in a slightly rewritten form):
$f(y)=f(x)+(y-x)f\prime(x)+\frac{1}{2}(y-x)^2 f\prime\prime(\theta)$
now plug in y=2 and y=0:
$f(0)=f(x)-xf\prime(x)+\frac{1}{2}x^2 f\prime\prime(\theta_1)$
$f(2)=f(x)+(2-x)f\prime(x)+\frac{1}{2}(2-x)^2 f\prime\prime(\theta_2)$
from first equation you can get e.g.
$xf\prime(x) =f(0)+f(x)+ \frac{1}{2}x^2 f\prime\prime(\theta_1) \le 1+1+2=4$
and second
$(2-x)f\prime(x) =f(2)+f(x)+ \frac{1}{2}(y-x)^2 f\prime\prime(\theta_2) \le 1+1+2=4$
add them up and you get $2f\prime(x) \le 8$, i.e. $f\prime(x) \le 4$. maybe i'm messing smth up, dunno... :/
• Apr 6th 2011, 06:48 AM
Opalg
Quote:

Originally Posted by choovuck
Taylor's expansion about x is (in a slightly rewritten form):
$f(y)=f(x)+(y-x)f'(x)+\frac{1}{2}(y-x)^2 f\prime\prime(\theta)$
now plug in y=2 and y=0:
$f(0)=f(x)-xf'(x)+\frac{1}{2}x^2 f''(\theta_1)$
$f(2)=f(x)+(2-x)f'(x)+\frac{1}{2}(2-x)^2 f''(\theta_2)$

This is an infuriating problem. I have come across variants of it many times, and it always takes me ages to get the answer to come out right. You need to do the steps of the argument in exactly the right order.

From the first equation,
$xf'(x) = f(x) - f(0) + \frac12x^2f''(\theta_1)$.

From the second equation,
$(2-x)f'(x) = f(2) - f(x) - \frac12(2-x)^2f''(\theta_2)$.

$2f'(x) = f(2) - f(0) + \frac12\bigl(x^2f''(\theta_1) - (2-x)^2f''(\theta_2)\bigr)$.

Therefore
$|f'(x)| \leqslant \frac12|f(2) - f(0)| + \frac14\bigl|x^2f''(\theta_1) - (2-x)^2f''(\theta_2)\bigr|$.

But $|f(2) - f(0)| \leqslant |f(2)| + |f(0)| \leqslant2$, and

$\bigl|x^2f''(\theta_1) - (2-x)^2f''(\theta_2)\bigr| \leqslant x^2 + (2-x)^2 = 4 - 2x(2-x)\leqslant4$.

Thus $|f'(x)| \leqslant 1+1=2$.